Euclidian topology ang cofinite topology

[lindita]

please can you help me to prove this exercise;
Prove that:
the Euclidean topology R is finer than the cofinite topology on R



please answer me as faster as u can I have an exame on monday and I don't know to provethis exercise!

 

[rasmhop]

Consider an arbitrary open set U in the cofinite topology. U is all of R except $x_1,x_2,\ldots,x_n$. Given some point $x \in U$ can you find an open interval I containing x, but included in U. That is $x \in I \subseteq U$ where I is open in the standard Euclidean topology and U is open in the cofinite topology. (HINT: can you show that the open interval centered at x with radius $\min(|x-x_1|,|x-x_2|,\ldots,|x-x_n|)$ works?)

It's a well-known theorem that if for every open set U in a topology T, and every point $x\in U$ we can find some set U' such that $x \in U' \subseteq U$ and U' is open in the topology T', then T' is finer than T.

 

[lindita]

Consider an arbitrary open set U in the cofinite topology. U is all of R except $x_1,x_2,\ldots,x_n$. Given some point $x \in U$ can you find an open interval I containing x, but included in U. That is $x \in I \subseteq U$ where I is open in the standard Euclidean topology and U is open in the cofinite topology. (HINT: can you show that the open interval centered at x with radius $\min(|x-x_1|,|x-x_2|,\ldots,|x-x_n|)$ works?)

It's a well-known theorem that if for every open set U in a topology T, and every point $x\in U$ we can find some set U' such that $x \in U' \subseteq U$ and U' is open in the topology T', then T' is finer than T.

thank you! U had helped me a lot thanks thanks

 

[lindita]

Can you help me and for this exercise please!

the function f:(X,T)->(X,T') is continuous
the function f:(X,T')->(X,T) is open



{T is the Euclidean topology,T' is cofinite topology}
[the function f maybe is the identity function id but I am not sure]

 

[rasmhop]

Can you help me and for this exercise please!

the function f:(X,T)->(X,T') is continuous
the function f:(X,T')->(X,T) is open



{T is the Euclidean topology,T' is cofinite topology}
[the function f maybe is the identity function id but I am not sure]

I'm assuming that you're asked to prove the two statements.

If f is the identity function then you're just restating the result in the OP. That f:(X,T)->(X,T') is continuous means that for every set U open in T', the set $f^{-1}(U) = U$ is open in T. Hence this is equivalent to "if U is open in T', then U is open in T" which is just stating that T is finer than T'. The other statement is shown in exactly the same way.

If f is an arbitrary function then the result is not true because for every $y \in X$ we can let $f_y : (X,T') \to (X,T)$ be the constant function f(x) = y so $f(X) = \{y\}$ is open according to your second statement. Since every one-element set is open T is the discrete topology, but the Euclidean topology is different from the discrete topology (unless of course X consists solely of isolated points).

 

[lindita]

Thank you!